Sync LeetCode submission Runtime - 1361 ms (76.23%), Memory - 25.9 MB (23.77%)

Author: jimit105

3491-find-the-maximum-length-of-valid-subsequence-ii/README.md

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+You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>.
+<p>A <span data-keyword="subsequence-array">subsequence</span> <code>sub</code> of <code>nums</code> with length <code>x</code> is called <strong>valid</strong> if it satisfies:</p>
+
+<ul>
+	<li><code>(sub[0] + sub[1]) % k == (sub[1] + sub[2]) % k == ... == (sub[x - 2] + sub[x - 1]) % k.</code></li>
+</ul>
+Return the length of the <strong>longest</strong> <strong>valid</strong> subsequence of <code>nums</code>.
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,5], k = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The longest valid subsequence is <code>[1, 2, 3, 4, 5]</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,4,2,3,1,4], k = 3</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The longest valid subsequence is <code>[1, 4, 1, 4]</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 10<sup>3</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>3</sup></code></li>
+</ul>

3491-find-the-maximum-length-of-valid-subsequence-ii/solution.py

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+# Approach: Dynamic Programming
+
+# Time: O(k^2 + n * k)
+# Space: O(k^2)
+
+class Solution:
+    def maximumLength(self, nums: List[int], k: int) -> int:
+        dp = [[0] * k for _ in range(k)] # dp[i][j] represents the length of sequence ending with remainder j where the previous remainder was i
+        res = 0
+
+        for num in nums:
+            num %= k
+            for prev in range(k):
+                dp[prev][num] = dp[num][prev] + 1
+                res = max(res, dp[prev][num])
+        return res
+